From the Hilbert's Nullstellensatz, we have the following consequence which is usually called Combinatorial Nullstellensatz:
Let $F$ be an arbitrary field, and let $f = f(x_1,x_2,\cdots,x_n)$ be a polynomial in $F(x_1,x_2,\cdots,x_n)$. Suppose the degree $deg(f)$ of $f$ is $\sum_{i=1}^nt_i$, where each $t_i$ is a nonnegative integer, and suppose the coecient of $\prod_{i=1}^nx_i^{t_i}$ in $f$ is nonzero. Then, if $S_1,S_2,\cdots,S_n$ are subsets of $F$ with $|S_i| > t_i$, there are $s_1 \in S_1; s_2 \in S_2; \cdots; s_n\in S_n$ so that $f(x_1,x_2,\cdots,x_n)\neq 0$.
But it is not right for the following inverse proposition：
Suppose the nonnegative integers $m_1,m_2,\cdots,m_n$ satisfies:
For any $S_1,S_2,\cdots,S_n\subset F$ with $|S_i|=m_i$, there are $s_1 \in S_1; s_2 \in S_2; \cdots; s_n\in S_n$ such that $f(x_1,x_2,\cdots,x_n)\neq 0$.
Then there exists a monomial $\prod_{i=1}^nx_i^{t_i}$ in $f$ such that $t_i<m_1,t_2<m_2,\cdots,t_n<m_n$.
For example, let $F$ be the real field $R$, $n=1$ and $f(x)=x^2$, For any $S\subset F$ with $|S|=2$, there is $s\in S$ such that $f(s)\neq 0$. But there does not exist a monomial in $f$ whose degree is $1$.
I want to ask whether I can add some stronger condition such that the following specific case is right:
Let $F$ be an infinite field, and let $f_i(x_1,x_2,\cdots,x_n),i=1,2,\cdots,k(\leq n)$ be polynomials in $F(x_1,x_2,\cdots,x_n)$ satisfy:
$(1)f_i\neq f_j,\forall1\leq i<j\leq n;$
$(2)deg(f_i)=1,\forall1\leq i\leq n;$
$(3)$For any $S_1,S_2,\cdots,S_n\subset F$ with $|S_i|=2$, there are $s_1 \in S_1; s_2 \in S_2; \cdots; s_n\in S_n$ such that $\prod_{i=1}^kf_i(s_1,s_2,\cdots,s_n)\neq 0;$.
$(4)$additional conition.
Let $f = f(x_1,x_2,\cdots,x_n)=\prod_{i=1}^kf_i(x_1,x_2,\cdots,x_n)$, then there must exist a monomial $x_{i_1}x_{i_2}\cdots x_{i_k}$ in $f$ such that $i_p\neq i_q,\forall 1\leq p<q\leq n.$
